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If `(sum_(n)^(n=1) (n^(2)+3n+3)(n+1)!)/((n+2)!)=8`, then n is equal to : |
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Answer» Correct Answer - 6 `because" "underset(n=-1)overset(n)(Sigma)(n^(2)+3n+3)(n+1)! = 8(n+2)!` `impliesunderset(n=-1)overset(n)(Sigma)((n+2)^(2)-(n+1)(n+1)! = 8 (n+2)!` `impliesunderset(n=-1)overset(n)(Sigma)((n+2).(n+2)!-(n+1)(n+1)!)=8(n+2)!` `implies(n+2).(n+2) ! = 8 (n+2)!` `implies n+2=8impliesn=6` |
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